Why single diode RF rectifier can realize above 50% power efficiency?

[lfx0828]

Hi guys,

Recently I have studied some RF-rectifier papers.
As the topic said, I am curious about the efficiency of single-diode RF-rectifiers.
I think it seems like a half-circle rectifier circuit which means the efficiency should less than 50%.
But many papers said the efficiency can up to 70% even 90%.
So, is there anyone can give me an explanation about this?
Thanks in advance.

 

[stenzer]

Hi,

please link the articles you are referring to. The upper figure does not indicate an efficiency larger than 50 %, and the lower one includes a matching network and does not mention a half wave rectifier (might be a simplified sketch).

BR

 

[lfx0828]

Hi,

please link the articles you are referring to. The upper figure does not indicate an efficiency larger than 50 %, and the lower one includes a matching network and does not mention a half wave rectifier (might be a simplified sketch).

BR

Thanks for your response.

I have uploaded the two related papers.

Looking forward to your response.

 

[FvM]

Your figure 5 is missing a rf choke similar to figure 3a in the first paper, DC block not necessarily required. Then it can achieve the indicated efficiency numbers.

 

[lfx0828]

Your figure 5 is missing a rf choke similar to figure 3a in the first paper, DC block not necessarily required. Then it can achieve the indicated efficiency numbers.

Thanks for your response.
But it seems I didn't describe my question clearly enough.
As you know, for this diode only the positive circle of the RF signal can pass through, which means only 50% power (maximum case) can be converted.
But in many papers, they said they can realize 70% even 80% efficiency. So, where is the other power comes from?
Or how is the signal transferred in the circuit?
For ease of description, the following figures show a simple simulation in PSpice.
Looking forward to your response.

 

[FvM]

What is the idea behind your circuit? You seem to struggle hard to reduce the efficiency. Better refer to the schematics in the quoted paper.

 

[vfone]

The efficiency of a half-wave rectifier is maximum 40.6%

AC Rectifier Efficiency

A rectifier is the device used to convert an AC signal into a DC signal. Half-wave rectifiers and full-wave rectifiers are used to do this conversion. This article explains how these rectifiers work, which rectifier is more effective in converting AC to DC, and how to justify computations...

www.brighthubengineering.com

How much efficient is half wave rectifier: explained with equation

analyseameter.com

 

[FvM]

The efficiency of a half-wave rectifier is maximum 40.6%

How? Efficieny is output power/ input power. No input power is consumed during negative halfwave. The circuit is however inappropiate to analyze RF rectifier efficiency.

 

[vfone]

40.6% is an theorethical number. The equations in the links above shows how to arrive to this number.
Is the same as when mentioned, the Class-B power amplifier has a maximum theoretical efficiency of ≈ 78.5%.

 

[FvM]

Sorry, the quoted equations seem completely wrong.

 

[mtwieg]

Thanks for your response.

I have uploaded the two related papers.

Looking forward to your response.

I think your question might be due to your own definition of efficiency. I'm guessing you are thinking of efficiency as η₁=P_DC/P_available. For such a definition, I would agree with your thinking.

However, the second paper defines efficiency as η₂=P_DC/(P_DC+P_Loss), which is entirely different. P_DC+P_Loss is basically the total absorbed power by the circuit, which would only be equal to P_available in the case of perfect conjugate matching (no reflected power).

The first efficiency η₁ is useful when you want to maximize output power. The second definition η₂ is useful when you want to minimize dissipated power. IMO η₁ is actually much more appropriate for energy harvesting applications like this. I doubt heating in the diode is a concern when harvesting microwatts of RF...

That aside, I don't see a way for η₁>0.5 with a half wave rectifier. But I also wouldn't rule out the possibility...

 

[FvM]

I have set up an ideal rectifier circuit to evaluate efficiency

By using an ideal diode, the circuit has no internal losses. I'm using a source with real impedance, it can be considered as equivalent of a generator or an antenna. Efficiency has to be related to the maximal delivered power of the source, in this case achieved with RL = 1k. Load voltage V2 is varied to for maximal output power which can be measured as average of V(p). The output efficiency is lower than the delivered source power because harmonic and possibly DC power is reflected back to the source.

Without a choke as DC path, we achieve an efficiency of about 46%, it's mainly limited by the load DC voltage drop. If you restrict the halfwave rectifier scope to this circuit class, efficiency is in fact lower than 50 %. The halfwave rectifiers in literature however are providing a DC path and optionally impedance matching means that reduce reflected harmonic power, e.g. the circuits discussed in paper 1 quoted in post #3. The simple DC path increases ideal rectifier efficiency to 81 %, harmonic filters can increase it further.

 

[lfx0828]

I have set up an ideal rectifier circuit to evaluate efficiency

View attachment 168325

By using an ideal diode, the circuit has no internal losses. I'm using a source with real impedance, it can be considered as equivalent of a generator or an antenna. Efficiency has to be related to the maximal delivered power of the source, in this case achieved with RL = 1k. Load voltage V2 is varied to for maximal output power which can be measured as average of V(p). The output efficiency is lower than the delivered source power because harmonic and possibly DC power is reflected back to the source.

Without a choke as DC path, we achieve an efficiency of about 46%, it's mainly limited by the load DC voltage drop. If you restrict the halfwave rectifier scope to this circuit class, efficiency is in fact lower than 50 %. The halfwave rectifiers in literature however are providing a DC path and optionally impedance matching means that reduce reflected harmonic power, e.g. the circuits discussed in paper 1 quoted in post #3. The simple DC path increases ideal rectifier efficiency to 81 %, harmonic filters can increase it further.

Thank you for such a good explanation.
Could you please tell me the software you used? It seems very good.

--- Updated Mar 27, 2021 ---

I think your question might be due to your own definition of efficiency. I'm guessing you are thinking of efficiency as η₁=P_DC/P_available. For such a definition, I would agree with your thinking.

However, the second paper defines efficiency as η₂=P_DC/(P_DC+P_Loss), which is entirely different. P_DC+P_Loss is basically the total absorbed power by the circuit, which would only be equal to P_available in the case of perfect conjugate matching (no reflected power).

The first efficiency η₁ is useful when you want to maximize output power. The second definition η₂ is useful when you want to minimize dissipated power. IMO η₁ is actually much more appropriate for energy harvesting applications like this. I doubt heating in the diode is a concern when harvesting microwatts of RF...

That aside, I don't see a way for η₁>0.5 with a half wave rectifier. But I also wouldn't rule out the possibility...

Thank you, yes you are right! That's what I am thinking about.

 

[vfone]

Sorry, the quoted equations seem completely wrong.

OK, they are wrong...what about this one?

Half Wave Rectifier Calculator

 

[FvM]

The term rectifier efficiency used by this calculator (as well as in the post #7 links) is misleading. It's said

The ratio of DC power obtained at the output to the applied input AC power is known as rectifier efficiency.

What they mean is not applied input power, it's apparent input power Vrms*Irms. That's surely an interesting quantity, but it's not directly related to the delivered power of a source with impedance.

 

[vfone]

Is well known that the detector input impedance is very important getting a good detector efficiency, but let's assume that the AC source is perfect matched to the detector input.
In your opinion, what other meanings will make the maximum efficiency number to be different from all my links above.

 

[FvM]

The commonly understood meaning of efficiency is real output power/ real input power. The AC sources in post #7 and #14 are ideal voltage sources and can't absorb reflected power. A halfwave rectifier with zero forward voltage (no internal losses) has respectively an efficiency of 1.0.